Beating the parity obstruction for primes p with μ(p − 1) = 1

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Beating the parity obstruction for primes p with μ(p − 1) = 1

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Take a shifted correlation with a sign twist in it and decide whether the sign survives the argument. That check is how the paper gets parity information past a sieve that can't see it, and it proves that infinitely many primes p have p − 1 squarefree with an even number of prime factors, that is, μ(p − 1) = 1.

Why the sieve is blind, and the weight that fixes it

Siegel–Walfisz, the large sieve and divisor weights all supply congruence information, and congruence information alone need not distinguish the two Liouville signs. This is the parity obstruction [0:17]. The paper writes p = 2u + 1, so Ω(p − 1) = 1 + Ω(u). It then builds a nonnegative weight A(u) = A₀(u)·(1 − E(u))/2, where E(u) is a Liouville-type sign that counts only primes from designated groups [1:03]. Group primes in u come with no square factors, so the ordinary slots carry a fixed, even number of primes and the projection keeps only an odd group count. Wherever A(u) > 0, Ω(u) is odd and Ω(2u) is even. In the toy case at [1:58], two slot primes times three group primes give E = −1 and Ω(p − 1) = 6.

The step that keeps the sign: ε(D_R)² = 1

Detecting primes needs Type II (bilinear) distribution, and that comes from a shifted correlation with the same sign ε at both endpoints and a prime in the long slot [2:30]. The graph argument writes n = D_R·w and n′ = D_R·w′. Since ε is completely multiplicative, ε(D_R w)ε(D_R w′) = ε(D_R)²ε(w)ε(w′) = ε(w)ε(w′). This identity is exact even when shared primes also divide w (eq. 4.18) [2:58]. The Möbius function fails here: μ(q·q) = 0, while μ(q)² = 1. A sign on only one endpoint also fails, because ε(D_R) then appears just once [3:41].

The rest of the proof

The prime slot needs cancellation in a prime Dirichlet polynomial at large heights. The frequencies split three ways: a small-prime factor confines the difficult ones to a sparse set, the new prime estimate takes the large ones, and a character–Mellin discrepancy takes the small ones [4:13]. Type II plus congruence distribution, bounded proxies and an upper bound for balanced semiprimes then leave positive prime mass [4:38].

The check to carry: is the sign completely multiplicative, and does it sit on both endpoints, so the shared dilation's sign appears squared? Use it when you read Section 4 [5:04].

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